I probably would first write the equation $y^{3/2}=5y$ as $y(y^{1/2})=5y$. This has the solution $y=0$. If $y \ne 0$, we can cancel $y$ from both sides (that is, divide both sides by $y$) and obtain $y^{1/2}=5$. Now square both sides. We get $y=25$.

So the two solutions of the equation are $y=0$ and $y=25$.

Remark: Let us explore your idea. You wanted to get a $y$ on the left-hand side by raising $y^{3/2}$ to a certain power. That can be done, but the appropriate power is $2/3$.

We have $(y^{3/2})^{2/3}=y^{(3/2)(2/3)}=y^1=y$. We need to do the same thing to the right-hand side, and we obtain the equation $y=5^{2/3}y^{2/3}$. Improved the left-hand side, but the right-hand side has been uglified.